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Not Homework. Proving question

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that:

    [tex](\frac{10^{n}}{10^{n}-1})^{10^n} = e [/tex]

    as n approaches inf.

    2. Relevant equations

    It is rather obvious that i can let 10^{n} be yet another variable


    3. The attempt at a solution

    I proved by assuming binomial dist as Poisson dist (it is interesting for me to use this to prove a continued fraction)
     
  2. jcsd
  3. Oct 18, 2011 #2

    Mark44

    Staff: Mentor

    You can simplify by letting u = 10n and working with this limit:

    [tex]\lim_{u \to \infty}\left( \frac{u}{u - 1} \right)^u[/tex]

    The usual approach is to let y = (u/(u - 1))u, and then take the natural log of both sides.

    ln y = u ln(u/(u - 1)) = [ln(u/(u - 1))]/(1/u)

    Now take the limit of both sides. Note that the expression above is the limit of the log of what you want.
     
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